Haskell功力太弱了,刷一下HackerRank上面的Haskell题目(同时在读Haskell函数式编程入门第二版),都记录在这个博客里面。
链接
目标:满分
f :: Int -> [Int] -> [Int]
f n arr = -- Complete this functio
case arr of
(fir:aft) -> (helper n fir) ++ (f n aft)
_ -> []
helper :: Int -> Int -> [Int]
helper n fir
| n > 0 = fir : helper (n-1) fir
| n == 0 = []
-- This part handles the Input and Output and can be used as it is. Do not modify this part.
main :: IO ()
main = getContents >>=
mapM_ print. (\(n:arr) -> f n arr). map read. words
f :: Int -> [Int] -> [Int]
f n (fir:aft) --Fill up this function
| fir < n = fir : f n aft
| otherwise = f n aft
f n [] = []
-- The Input/Output section. You do not need to change or modify this part
main = do
n <- readLn :: IO Int
inputdata <- getContents
let
numbers = map read (lines inputdata) :: [Int]
putStrLn . unlines $ (map show . f n) numbers
f :: [Int] -> [Int]
f [] = []-- Fill up this Function
f (x : (y : ys)) = y : f ys
f (x : xs) = xs
-- This part deals with the Input and Output and can be used as it is. Do not modify it.
main = do
inputdata <- getContents
mapM_ (putStrLn. show). f. map read. lines $ inputdata
fn n
| n > 0 = 1 : fn (n-1)
| otherwise = []
main = do
n <- readLn :: IO Int
print (fn(n))
rev (x : xs) = (rev xs) ++ [x]
rev [] = []
这是个问答题...
((λx.(x y))(λz.z))
把x替换为(λz.z))
就是(λz.z)y
applicate y,得到y
((λx.((λy.(x y))x))(λz.w))
((λy.((λz.w) y))(λz.w))
反正往下最左结合,得到w
4
6
难度飙升
只需要值域和定义域一一对应即可,但是实现也不简单
值得注意的是
tuple仍然属于Ord的范畴,所以里面嵌套了一个快排的标准实现
quickSort :: Ord a => [a] -> [a]
quickSort [] = []
quickSort (x:xs) =
let smallSort = quickSort [a | a <- xs, a < x]
biggerSort = quickSort [a | a <- xs, a > x]
in smallSort ++ [x] ++ biggerSort
isFunc :: [(Int, Int)] -> Bool
isFunc [] = True
isFunc ((x1,x2):xs)
| xs == [] = True
| (x1 == (fst (head xs))) && (x2 /= (snd (head xs))) = False
| otherwise = isFunc xs
tuplify :: [Int] -> [(Int, Int)]
tuplify [] = []
tuplify (x:y:xs) = (x,y):(tuplify xs)
solveOne :: [Int] -> String
solveOne xs
| isFunc $ quickSort $ tuplify xs = "YES"
| otherwise = "NO"
-- splitAT 分割剩下的数组 splitAt :: Int -> [a] -> ([a], [a])
solve :: [Int] -> [String]
solve [] = []
solve (n:xs) = (solveOne f):(solve s)
where (f, s) = splitAt (2*n) xs
-- unlines [String] -> String 转换数组为一个String
-- words 则为其反运算 String -> [String]
-- tail 用于去掉不需要的输入
-- read 用于类型转换
-- interact 从标准输入读入,处理后从标准输出写入
main = interact $ unlines . solve . map read . tail . words
-- main = unlines . solve . map read . tail . words
f ::[Int] -> Int
f [] = 0
f arr -- Fill up this function body
| odd $ head $ arr = (head arr) + (f $ tail arr)
| otherwise = f $ tail arr
-- This part handles the Input/Output and can be used as it is. Do not change or modify it.
main = do
inputdata <- getContents
putStrLn $ show $ f $ map (read :: String -> Int) $ lines inputdata
len :: [a] -> Int
len lst = length lst
-- Enter your code here. Read input from STDIN. Print output to STDOUT
f [] = []
f (x:xs) -- Complete this function here
| x < 0 = (-x):(f xs)
| otherwise = x : (f xs)
-- This section handles the Input/Output and can be used as it is. Do not modify it.
main = do
inputdata <- getContents
mapM_ putStrLn $ map show $ f $ map (read :: String -> Int) $ lines inputdata
import Text.Printf (printf)
-- This function should return a list [area, volume].
solve :: Double -> Double -> [Double] -> [Double] -> [Double]
solve left right coeffs expnts = [area, volume]
where
step = 0.001
domain = [left, left+step..right]
bigF x = sum $ zipWith (\c e -> c*x**e) coeffs expnts
area = sum . map (*step) . map bigF $ domain
volume = sum . map (*step) . map (\r -> pi*r**2) . map bigF $ domain
--Input/Output.
main :: IO ()
main = getContents >>= mapM_ (printf "%.1f\n"). (\[a, b, [l, r]] -> solve l r a b). map (map read. words). lines
-- Enter your code here. Read input from STDIN. Print output to STDOUT
solvehandler (a:b:xs) = solve ([a,b] ++ xs ++ [a,b])
solve :: [Int] -> Double
solve [] = 0.0
solve [a,b] = 0.0
solve (a:b:c:d:xs) = (solve $ c:d:xs) + (sqrt $ fromIntegral $ x'*x' + y'*y')
where
x' = c - a
y' = d - b
main = interact $ show . solvehandler . map read . tail . words
多边形计算面积 —— 但是是函数式
本来想的是通过外面的四边形的面积减去每个小三角形,没想到这种只适合凸多边形。。。
-- 适合凸多边形的面积计算
-- Enter your code here. Read input from STDIN. Print output to STDOUT
import Data.List
import Data.Ord
solvehandler (a:b:xs) = solve ([a,b] ++ xs ++ [a,b])
abs a
| a > 0 = a
| otherwise = -a
intToTuple :: [Int] -> [(Int, Int)]
intToTuple [] = []
intToTuple [a,b] = [(a,b)]
intToTuple (a:b:c) = (a,b):(intToTuple c)
calcuteBlankArea :: [Int] -> Double
calcuteBlankArea [] = 0.0
calcuteBlankArea [a,b] = 0.0
calcuteBlankArea (a:b:c:d:xs) = (calcuteBlankArea $ c:d:xs) + (Main.abs $ fromIntegral $ (div (x*y) 2))
where
x = c - a
y = d - b
solve :: [Int] -> Double
solve [] = 0.0
solve ls = (fromIntegral total) - (calcuteBlankArea ls)
where
transTuple = intToTuple ls
minx = snd $ (maximumBy (comparing snd) transTuple)
maxx = snd $ minimumBy (comparing snd) transTuple
miny = fst $ maximumBy (comparing fst) transTuple
maxy = fst $ minimumBy (comparing fst) transTuple
total = ((maxx - minx) * (maxy - miny))
main = interact $ show . solvehandler . map read . tail . words
正确答案:
main = interact $ show . solve . lines
solve :: [String] -> Double
solve (x:xs) = abs . fst $ foldl accFun (0, last points) points
where
n = read x :: Int
points = map ((\[a,b] -> (read a, read b)) . words ) $ take n xs :: [(Double, Double)]
areaUnder (a,b) (c,d) = (a-c) * abs (b+d) / 2
accFun = \(accVal, lastPoint) currPoint -> (accVal + (areaUnder lastPoint currPoint), currPoint)
--Contributed by Ron Watkins
module Main where
fib n -- Enter your code here to complete this function
| n == 1 = 0
| n == 2 = 1
| otherwise = fib(n-1) + fib(n-2)
-- This part is related to the Input/Output and can be used as it is
-- Do not modify it
main = do
input <- getLine
print . fib . (read :: String -> Int) $ input
杨辉三角的下一行输出可以表示为
0:上一行
上一行:0
的加和
solve :: Int -> [[Int]]
solve 1 = [[1]]
solve 2 = solve 1 ++ [[1,1]]
solve n = solve (n-1) ++ [zipWith (+) (0:(last . solve $ (n-1))) ((last . solve $ (n-1)) ++ [0])]
main = interact $ unlines . map unwords . map (\xs -> [ show x | x <- xs ]) . solve . read
-- Enter your code here. Read input from STDIN. Print output to STDOUT
solve :: [String] -> [String]
solve [f,s] =
zipWith (\a b -> [a,b]) f s
main = interact $ filter (\a -> a/=' ') . unwords . solve . words
-- Enter your code here. Read input from STDIN. Print output to STDOUT
import Data.List
solve :: [String] -> String
solve [] = []
solve (x:xs)
| length x > 1 = [head x] ++ (show (length x)) ++ solve xs
| otherwise = [head x] ++ solve xs
main = interact $ solve . group
-- Enter your code here. Read input from STDIN. Print output to STDOUT
solve :: String -> String
solve [] = []
solve (x:y:xs) = [y,x] ++ solve xs
main = interact $ unlines . map solve . tail . words
showLen :: [Char] -> String
showLen = unwords . sequence [show . length, id]
prefixLength :: Eq a => ([a], [a]) -> Int
prefixLength = length . takeWhile id . uncurry (zipWith (==))
prefixCompression :: Eq a => [a] -> [a] -> ([a], [a], [a])
prefixCompression xs ys = let p = prefixLength (xs, ys) in (take p xs, drop p xs, drop p ys)
main = do
x <- getLine
y <- getLine
let (p, x', y') = prefixCompression x y in do
putStrLn $ showLen p
putStrLn $ showLen x'
putStrLn $ showLen y'
先是练习α β η 规约的几道题目(貌似没η)η就是point-free了
beta规约下式:
((λx.(x x))(λx.(x x)))
一眼顶针,鉴定为不动点
CAN'T REDUCE
规约这个
(λg.((λf.((λx.(f (x x)))(λx.(f (x x))))) g))
得到
λg.(((((g (λx.(g (x x) λx.(g (x x)))))))
一眼顶针,鉴定为不动点
CAN'T REDUCE
此题询问丘奇编码下下列数的值
λx.λy.x47y
这是个丘奇数表示的基本问题
Church数0, 1, 2, ...在lambda演算中被定义如下:
0 ≡ λf.λx. x
1 ≡ λf.λx. f x
2 ≡ λf.λx. f (f x)
3 ≡ λf.λx. f (f (f x))
所以这题是47
Compute the value of λx.λy.x(xy).
2
Compute the value of λx.λy.x(xy).
0
这几个题都是丘奇数的入门
f :: Int -> [Int] -> [Int]
f n arr = -- Complete this functio
case arr of
(fir:aft) -> (helper n fir) ++ (f n aft)
_ -> []
helper :: Int -> Int -> [Int]
helper n fir
| n > 0 = fir : helper (n-1) fir
| n == 0 = []
-- This part handles the Input and Output and can be used as it is. Do not modify this part.
main :: IO ()
main = getContents >>=
mapM_ print. (\(n:arr) -> f n arr). map read. words
f :: Int -> [Int] -> [Int]
f n (fir:aft) --Fill up this function
| fir < n = fir : f n aft
| otherwise = f n aft
f n [] = []
-- The Input/Output section. You do not need to change or modify this part
main = do
n <- readLn :: IO Int
inputdata <- getContents
let
numbers = map read (lines inputdata) :: [Int]
putStrLn . unlines $ (map show . f n) numbers
f :: [Int] -> [Int]
f [] = []-- Fill up this Function
f (x : (y : ys)) = y : f ys
f (x : xs) = xs
-- This part deals with the Input and Output and can be used as it is. Do not modify it.
main = do
inputdata <- getContents
mapM_ (putStrLn. show). f. map read. lines $ inputdata